Question

Cho M= 1-\(\frac{1}{2^2}\)-\(\frac{1}{3^2}\)-\(\frac{1}{4^2}\)-....-\(\frac{1}{100^2}\). CMR: M>\(\frac{1}{100}\)

Answer 1

M = \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}\)

M = 1 - (\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\))

Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\) = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(1-\frac{1}{100}\)

M > 1 - (1 - \(\frac{1}{100}\)) =\(\frac{1}{100}\) (đpcm)